Question

A triangle is formed by joining three points A (8, 2), B (-4, -4) and C (16, 1) on rectangular hyperbola xy = 16. If orthocentre of the triangle ABC is (h, k) then product hk is equal to
a) 4
b) 8
c) 16
d) 32

Answer 1

Hint: Use the fact that AH and BC are perpendicular and BH and CA are perpendicular thus find points of orthocentre.

Complete step-by-step answer:
In the question a triangle is given or formed using three points A (8, 2) B (-4, -4) and C (16, 1) on rectangular hyperbola xy = 16. So, if the orthocentre of triangle ABC is (h, k) then we have to find the product of h and k. Let’s draw the triangle

Now as we know that H is orthocentre so the product of slope of AH and BC will be – 1 as AH when extended it will be the altitude of the triangle from point A to BC. We will find slope AH and BC using formula,
$m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$
Where $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)$ are points. So, slope of AH where A = (8, 2) and H = (h, k) will be
$m=\dfrac{k-2}{h-8}$
And slope of BC where B = (-4, -4) and C = (16, 1) will be,
$m=\dfrac{1-\left( -4 \right)}{16-\left( -4 \right)}=\dfrac{5}{20}=\dfrac{1}{4}$
So, we can say that
$\dfrac{k-2}{h-8}\times \dfrac{+1}{4}=-1$
Now on cross multiplication we get,
k – 2 = - 4 (h – 8)
Now on simplification we get,
K – 2 = - 4h + 32
Hence, on rearranging we get,
4h + k = 34…………………(i)
We can write k = 34 – 4h. Similarly we can also say that the product of slopes of BH and AC is also – 1 as H is the orthocentre and BH when extended represents altitude from B to AC. So, the slope of BH will be if point B and H is (-4, -4) and (h. k) respectively,
$m=\dfrac{k-\left( -4 \right)}{h-\left( -4 \right)}=\dfrac{k+4}{h+4}$
And slope of AC where A and C is (8, 2) and (16, 1) respectively,
$m=\dfrac{2-1}{8-16}=\dfrac{-1}{8}$
So we can say that,
$\dfrac{k+4}{h+4}\times \dfrac{-1}{8}=-1$
So now on cross multiplication we get,
k + 4 = 8 (h + 4)
Now on simplification we get,
k + 4 = 8h + 32
So, K = 8h + 28
We also know that k = 34 - 4h so we can write that,
34 – 4h = 8h + 28
Hence on simplification,
12h = 6 or
$h=\dfrac{1}{2}$
As we know that k = 34 – 4h and $h=\dfrac{1}{2}$ so,
$k=34-4\times \dfrac{1}{2}=34-2=32$
So we know that $h=\dfrac{1}{2}$ and k = 32 so,
$hk=\dfrac{1}{2}\times 32=16$ .
So the correct option is (c).

Note: If someone draws the triangle on the graph paper can also find the orthocentre on it then he or she will find that the orthocentre point lies on hyperbola so they can say the answer directly.